Frequency Response of BJT Amplifiers

Analog & Digital Electronics Course No: PH-218

Lec-11: Frequency Response of BJT Amplifiers

Course Instructors: Dr. A. P. VAJPEYI

Department of Physics, Indian Institute of Technology Guwahati, India 1 Frequency Response of CE BJT Amplifier

1 X = C 2π × f ×C

Assuming that the coupling and bypass capacitors are ideal shorts at the midrange signal frequency, the midrange voltage gain can be determined by

(RC \\ RL ) Av,mid = ' re

In the low frequency range, BJT amplifier has three high-pass RC circuits, namely input, bypass and output RC circuit, that affect its gain.

The lower cutoff frequency of a given common emitter amplifier will be given by the highest of the individual RC circuits.

fC-low = MAX(f c-input , fc-output ,f c-bypass ) 2 Low Frequency Response of Input RC circuit

The input RC Circuit

R in = R1 \\ R2 \\ Rin −base R V = ( in )V B 2 2 in X c1 + Rin

As the signal frequency decreases, X C1 increase, This causes less voltage across the input resistance of the amplifier at the base and because of this, the overall voltage gain of the amplifier is reduced.

3 Decibel

Bel is a form of gain measurement and is commonly used to express amplifier response.

The Bel is a logarithm measurement of the ratio of one power to another or one voltage to another. G(dB ) =10 log 10 (P2 / P1) G = log 10 (P2 / P1) G(dB ) = 20 log 10 (V2 /V1)

It was found, that the Bel was too large a unit of measurement for practical purposes, so the decibel (dB) was defined such that 1B =10dB

0 dB reference It is often convenient in amplifiers to assign a certain value of gain as the 0 dB reference This does not mean that the actual voltage gain is 1 (which is 0 dB); it means that the reference gain, is used as a reference with which to compare other values of gain and is therefore assigned a 0 dB value. The maximum gain is called the midrange gain and is assigned a 0 dB value. Any value of gain below midrange can be referenced to 0 dB and expressed as a negative dB value. 4 Bode Plots A plot of dB voltage gain versus frequency on semilog graph paper is called a bode plot.

The Bode Plot is a variation of the basic frequency response curve. A Bode plot assumes the amplitude is zero until the cutoff frequency is reached. Then the gain of the amplifier is assumed to drop at a set rate of 20 dB/decade (or one RC time constant).

5 Low Frequency Response of Input RC ckt A critical point in the amplifier's response occurs when the output voltage is 70.7% of its midrange value. This condition occurs in the input RC circuit when X C1 = R in R In terms of measurement in decibels: V = ( in )V B 2 2 in 20 log( V /V ) = 20 log( .0 707 ) = −3dB X c1 + Rin out in

Lower critical frequency The condition where the gain is down 3 dB is called the -3dB point of

the amplifier response; The frequency f c at which the overall gain is 3dB less than at midrange is called the lower cutoff frequency . 1 1 X = = R fC = C1 in 2π ×(R + R )×C 2π × fc ×C1 s in 1

R R R Where R s is the signal internal resistance and R in = 1 \\ 2 \\ in −base 6 Voltage Gain Roll Off for input ckt at low frequency The input RC circuit reduces the overall voltage gain of an amplifier

by 3 dB when the frequency is reduced to the critical value fC.

As the frequency continues to decrease below fC the overall voltage gain also continues to decrease. The decrease in voltage gain with frequency is called roll-off.

For each ten times reduction in

frequency below f C there is a 20dB reduction in voltage gain.

At f c , XC1 = R in , so X C1 =10 R in at 0.1f c, V R B = in = 1.0 V 2 2 in X c1 + Rin

20 log( VB /Vin ) = 20 log( )1.0 = −20 dB

7 Phase shift for input RC ckt at low frequency

At lower frequencies, higher values of X C1 cause a phase shift to be introduced, and the output voltage leads the input voltage.

The phase angle in an input RC circuit is expressed as: X θ = tan −1( C1 ) Rin At midrange frequencies the phase shift through the input RC

circuit is zero because X C1 ≈ 0 Ω. θ

8 Output RC circuit at low frequency 1 1 X = = R + R f = C3 2π × f ×C C L C 2π ×(R + R )×C c 3 C L 3

As the signal frequency decreases, XC3 increases. This causes less voltage across the load resistance because more voltage is

dropped across C3. The signal voltage is reduced by a factor of 0.707 when frequency

is reduced to the lower critical value, fC, for the circuit. This corresponds to a 3 dB reduction in voltage gain 9 Phase shift for Output RC ckt at low frequency

The phase shift in the output RC circuit is

X θ = tan −1( C3 ) RC + RL

θ ≈ 0 for the midrange frequency and approaches 90º as the

frequency approaches zero (X C3 approaches infinity).

At the critical frequency f C , the phase shift is 45º

10 Emitter-bypass RC ckt at low frequency

1 f Ve ' C = R = + r ' R in −emitter e 2π ×[( r + th //) R ]×C Ie e E 2 βac R I R R = th b + r ' = th + r ' in −emitter βI e β e b 11 Total Low frequency Response of CE Amplifier The critical frequencies of the three RC circuits are not necessarily all equal. If one of the RC circuits has a critical frequency higher than the other two, then it is dominant RC circuit.

As the frequency is reduced from midrange, the first "break point" occurs at the critical frequency of the input RC circuit, f c(input) , and the gain begins to drop at -20dB/decade.

This constant roll \off rate continues until the critical frequency of the output RC circuit, f c(output), is reached. At this break point, the output RC circuit adds another - 20 dB/decade to make a total roll-off of -40 dB/decade.

This constant -40 dB/decade roll-off continues until the critical frequency of the bypass RC circuit, f c(bypass), is reached. At this break point, the bypass RC circuit adds still another -20dB/decade, making the gain roll-off at - 60 dB/decade

12 If all RC circuits have different critical frequency If all RC circuits have the same critical frequency

http://www.uotiq.org/dep-eee/lectures/2nd/Electronics%201/part6.pdf

13 Low frequency Response of CE Amplifier Determine the value of the lower cutoff frequency for the following amplifier. Consider the following component values:

RS = 600 Ω, R1 = 18 k Ω, R2 = 4.7 k Ω, RC = 1.5 k Ω, RE = 1.2 k Ω, RL = 5 k Ω, CC1 = 1 µF, C C2 = 0.22 µF, C E = 10 µF, hfe = 200, hie = 4.4 k Ω, Vcc = 10 V

fc-input = 61.2 Hz fc-output = 111 Hz fc-bypass = 650 Hz

Lower cutoff frequency of the amplifier:

fC-Low = max (61.2, 111, 650) fC-Low = 650 Hz